Find the x and y components of the following vectors a a 10...

Find the x and y components of the following vectors (a) A 10-m displacement vector that makes an angle of 30? with the +x direction. (b) A 20-m/s velocity vector that makes an angle of 37? counterclockwise from the ?x direction. (c) A 80-N force vector that makes an angle of135? counterclockwise from the ?y direction. (d) The X and Y components of a force F acting at 30 degrees to x-axis are respectively. (e) A force of 19 N is in the direction of the negative x axis

Transcript

00:01 Okay, so for this question, we've got kind of a rapid fire getting x and y components from vectors.

00:08 For all of these, i'm pretty much going to be using sokatoa to find the components.

00:13 So i just kind of have the definitions in case you don't know them at the top for defining sine, cosine, and tangent with respect to the different sides of the right triangle that we'll make when we break down these vectors.

00:26 So the first one that we have is a 10 meter displacement going at 30 degrees with respect to the positive x -axis.

00:37 So we would have an x component and a y component.

00:45 And then based on the definitions that we have here, x is going to be the adjacent side with respect to that 30 degrees.

00:52 So we have adjacent over the hypotenuse, which is equal to 10 meters.

00:57 Is going to be equal to cosine of 30 degrees.

01:01 And then we can multiply each side by that 10 meters.

01:08 So we'll get x is 10 meters times cosine of 30 degrees, which when i plug that into my calculator, i get 8 .66 meters.

01:28 And then following that logic then for the y component we'll have y which is the opposite side over hypotenuse 10 meters is going to be equal to the sign of 30 and again we'll multiply up the 10 meters to get y by itself so the y component will be 10 meters times sign 30 which will be equal to 5 meters.

01:59 Then based on the, there was a little bit hard to tell from the way that the question was typed up, but i think that this velocity vector for part b was supposed to be an angle of 37 degrees counterclockwise from the negative x direction.

02:17 So i'll also break this up.

02:19 We'll call them vx and vy.

02:26 I'd say one of the disadvantages to using the sokatoa method is that you do have to use the text.

02:32 To kind of figure out what the signs are for the x and y component.

02:38 So we can see from the picture i drew here for part b, the vertical component we're going to have to just put a negative sign in front based on the context of the problem.

02:47 But otherwise, we're doing the same thing.

02:49 So we'll have vx over 20 is equal to.

02:55 So again, that's the adjacent side.

02:57 So that'll be cosine 37.

03:00 So vx is going to be equal to 20.

03:02 Times cosine of 37 which is equal to about 16.

03:11 I'm going to round to two sig figs...